# How do you differentiate x^4(x+y)=y^2(3x-y)?

Jun 1, 2015

First, use the distributive property to write ${x}^{5} + {x}^{4} y = 3 x {y}^{2} - {y}^{3}$.

Next, make the assumption that $y$ is a function of $x$ and differentiate both sides with respect to $x$, using the Product Rule and Chain Rule as necessary:

$5 {x}^{4} + 4 {x}^{3} y + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {y}^{2} + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

Finally, solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{4} - 6 x y + 3 {y}^{2}\right) = - 5 {x}^{4} - 4 {x}^{3} y + 3 {y}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{- 5 {x}^{4} - 4 {x}^{3} y + 3 {y}^{2}}{{x}^{4} - 6 x y + 3 {y}^{2}}$